The heart rates in beats per minute for runners who walk briskly on a treadmill for 6 minutes follow a Normal distribution with a mean of 104 and standard deviation of 12.5. What percent of the runners have heart rates above 140?

0.2%

Step-by-step explanation:

We apply the z score formula.

This is given as:

Z score = x - μ/σ

x = Raw score = 140

μ = Population mean = 104

σ = Standard deviation = 12.5

z = 140 - 104/12.5

z = 2.88

Probability value from Z-Table:

P(x<140) = 0.99801

P(x>140) = 1 - P(x<140)

= 1 - 0.99801

= 0.0019884

Approximately = 0.0020

Converting to percent = 0.2%

Therefore, percent of the runners have heart rates above 140 is 0.2%

you didn't give me the choices, so i will just give you the graph so you can look at your choices and choose the right one.

the answer is c.

step-by-step explanation: